例子:

absolute="/foo/bar"
current="/foo/baz/foo"

# Magic

relative="../../bar"

我如何创造魔法(希望不是太复杂的代码…)?


当前回答

我把你的问题作为一个挑战,用“可移植的”shell代码来编写它,即。

考虑到POSIX外壳 没有数组之类的bashisms 避免像打瘟疫一样打外部电话。脚本中没有一个分叉!这使得它非常快,特别是在有显著分叉开销的系统上,比如cygwin。 必须处理路径名中的glob字符(*,?,[,])

它运行在任何POSIX兼容shell (zsh, bash, ksh, ash, busybox,…)上。它甚至包含一个测试套件来验证其操作。路径名的规范化留作练习。: -)

#!/bin/sh

# Find common parent directory path for a pair of paths.
# Call with two pathnames as args, e.g.
# commondirpart foo/bar foo/baz/bat -> result="foo/"
# The result is either empty or ends with "/".
commondirpart () {
   result=""
   while test ${#1} -gt 0 -a ${#2} -gt 0; do
      if test "${1%${1#?}}" != "${2%${2#?}}"; then   # First characters the same?
         break                                       # No, we're done comparing.
      fi
      result="$result${1%${1#?}}"                    # Yes, append to result.
      set -- "${1#?}" "${2#?}"                       # Chop first char off both strings.
   done
   case "$result" in
   (""|*/) ;;
   (*)     result="${result%/*}/";;
   esac
}

# Turn foo/bar/baz into ../../..
#
dir2dotdot () {
   OLDIFS="$IFS" IFS="/" result=""
   for dir in $1; do
      result="$result../"
   done
   result="${result%/}"
   IFS="$OLDIFS"
}

# Call with FROM TO args.
relativepath () {
   case "$1" in
   (*//*|*/./*|*/../*|*?/|*/.|*/..)
      printf '%s\n' "'$1' not canonical"; exit 1;;
   (/*)
      from="${1#?}";;
   (*)
      printf '%s\n' "'$1' not absolute"; exit 1;;
   esac
   case "$2" in
   (*//*|*/./*|*/../*|*?/|*/.|*/..)
      printf '%s\n' "'$2' not canonical"; exit 1;;
   (/*)
      to="${2#?}";;
   (*)
      printf '%s\n' "'$2' not absolute"; exit 1;;
   esac

   case "$to" in
   ("$from")   # Identical directories.
      result=".";;
   ("$from"/*) # From /x to /x/foo/bar -> foo/bar
      result="${to##$from/}";;
   ("")        # From /foo/bar to / -> ../..
      dir2dotdot "$from";;
   (*)
      case "$from" in
      ("$to"/*)       # From /x/foo/bar to /x -> ../..
         dir2dotdot "${from##$to/}";;
      (*)             # Everything else.
         commondirpart "$from" "$to"
         common="$result"
         dir2dotdot "${from#$common}"
         result="$result/${to#$common}"
      esac
      ;;
   esac
}

set -f # noglob

set -x
cat <<EOF |
/ / .
/- /- .
/? /? .
/?? /?? .
/??? /??? .
/?* /?* .
/* /* .
/* /** ../**
/* /*** ../***
/*.* /*.** ../*.**
/*.??? /*.?? ../*.??
/[] /[] .
/[a-z]* /[0-9]* ../[0-9]*
/foo /foo .
/foo / ..
/foo/bar / ../..
/foo/bar /foo ..
/foo/bar /foo/baz ../baz
/foo/bar /bar/foo  ../../bar/foo
/foo/bar/baz /gnarf/blurfl/blubb ../../../gnarf/blurfl/blubb
/foo/bar/baz /gnarf ../../../gnarf
/foo/bar/baz /foo/baz ../../baz
/foo. /bar. ../bar.
EOF
while read FROM TO VIA; do
   relativepath "$FROM" "$TO"
   printf '%s\n' "FROM: $FROM" "TO:   $TO" "VIA:  $result"
   if test "$result" != "$VIA"; then
      printf '%s\n' "OOOPS! Expected '$VIA' but got '$result'"
   fi
done

# vi: set tabstop=3 shiftwidth=3 expandtab fileformat=unix :

其他回答

我把你的问题作为一个挑战,用“可移植的”shell代码来编写它,即。

考虑到POSIX外壳 没有数组之类的bashisms 避免像打瘟疫一样打外部电话。脚本中没有一个分叉!这使得它非常快,特别是在有显著分叉开销的系统上,比如cygwin。 必须处理路径名中的glob字符(*,?,[,])

它运行在任何POSIX兼容shell (zsh, bash, ksh, ash, busybox,…)上。它甚至包含一个测试套件来验证其操作。路径名的规范化留作练习。: -)

#!/bin/sh

# Find common parent directory path for a pair of paths.
# Call with two pathnames as args, e.g.
# commondirpart foo/bar foo/baz/bat -> result="foo/"
# The result is either empty or ends with "/".
commondirpart () {
   result=""
   while test ${#1} -gt 0 -a ${#2} -gt 0; do
      if test "${1%${1#?}}" != "${2%${2#?}}"; then   # First characters the same?
         break                                       # No, we're done comparing.
      fi
      result="$result${1%${1#?}}"                    # Yes, append to result.
      set -- "${1#?}" "${2#?}"                       # Chop first char off both strings.
   done
   case "$result" in
   (""|*/) ;;
   (*)     result="${result%/*}/";;
   esac
}

# Turn foo/bar/baz into ../../..
#
dir2dotdot () {
   OLDIFS="$IFS" IFS="/" result=""
   for dir in $1; do
      result="$result../"
   done
   result="${result%/}"
   IFS="$OLDIFS"
}

# Call with FROM TO args.
relativepath () {
   case "$1" in
   (*//*|*/./*|*/../*|*?/|*/.|*/..)
      printf '%s\n' "'$1' not canonical"; exit 1;;
   (/*)
      from="${1#?}";;
   (*)
      printf '%s\n' "'$1' not absolute"; exit 1;;
   esac
   case "$2" in
   (*//*|*/./*|*/../*|*?/|*/.|*/..)
      printf '%s\n' "'$2' not canonical"; exit 1;;
   (/*)
      to="${2#?}";;
   (*)
      printf '%s\n' "'$2' not absolute"; exit 1;;
   esac

   case "$to" in
   ("$from")   # Identical directories.
      result=".";;
   ("$from"/*) # From /x to /x/foo/bar -> foo/bar
      result="${to##$from/}";;
   ("")        # From /foo/bar to / -> ../..
      dir2dotdot "$from";;
   (*)
      case "$from" in
      ("$to"/*)       # From /x/foo/bar to /x -> ../..
         dir2dotdot "${from##$to/}";;
      (*)             # Everything else.
         commondirpart "$from" "$to"
         common="$result"
         dir2dotdot "${from#$common}"
         result="$result/${to#$common}"
      esac
      ;;
   esac
}

set -f # noglob

set -x
cat <<EOF |
/ / .
/- /- .
/? /? .
/?? /?? .
/??? /??? .
/?* /?* .
/* /* .
/* /** ../**
/* /*** ../***
/*.* /*.** ../*.**
/*.??? /*.?? ../*.??
/[] /[] .
/[a-z]* /[0-9]* ../[0-9]*
/foo /foo .
/foo / ..
/foo/bar / ../..
/foo/bar /foo ..
/foo/bar /foo/baz ../baz
/foo/bar /bar/foo  ../../bar/foo
/foo/bar/baz /gnarf/blurfl/blubb ../../../gnarf/blurfl/blubb
/foo/bar/baz /gnarf ../../../gnarf
/foo/bar/baz /foo/baz ../../baz
/foo. /bar. ../bar.
EOF
while read FROM TO VIA; do
   relativepath "$FROM" "$TO"
   printf '%s\n' "FROM: $FROM" "TO:   $TO" "VIA:  $result"
   if test "$result" != "$VIA"; then
      printf '%s\n' "OOOPS! Expected '$VIA' but got '$result'"
   fi
done

# vi: set tabstop=3 shiftwidth=3 expandtab fileformat=unix :

Python的os.path.relpath作为shell函数

这个relpath练习的目标是模仿xni提出的Python 2.7的os.path.relpath函数(从Python 2.6版可用,但只能在2.7中正常工作)。因此,一些结果可能与其他答案中提供的函数不同。

(我没有在路径中测试换行符,因为它破坏了基于从ZSH调用python -c的验证。经过一些努力,这当然是可能的。)

关于Bash中的“魔法”,我很久以前就放弃了在Bash中寻找魔法,但我已经在ZSH中找到了我需要的所有魔法,然后是一些。

因此,我提出了两种实现。

第一个实现的目标是完全兼容posix。我已经在Debian 6.0.6的“挤压”上用/bin/dash测试了它。它还可以在OS X 10.8.3上完美地与/bin/sh一起工作,这实际上是伪装成POSIX shell的Bash版本3.2。

第二个实现是一个ZSH shell函数,它对路径中的多个斜杠和其他麻烦具有健壮性。如果您有可用的ZSH,这是推荐的版本,即使您是在下面给出的脚本形式中调用它(即使用#!/usr/bin/env zsh)。

最后,我编写了一个ZSH脚本,根据其他答案中提供的测试用例,验证$PATH中relpath命令的输出。我通过添加一些空格、制表符和标点符号,例如!? *在这里和那里,还抛出了另一个测试与vim-powerline中发现的奇异UTF-8字符。

POSIX外壳函数

首先,posix兼容的shell函数。它适用于各种路径,但不清除多个斜杠或解析符号链接。

#!/bin/sh
relpath () {
    [ $# -ge 1 ] && [ $# -le 2 ] || return 1
    current="${2:+"$1"}"
    target="${2:-"$1"}"
    [ "$target" != . ] || target=/
    target="/${target##/}"
    [ "$current" != . ] || current=/
    current="${current:="/"}"
    current="/${current##/}"
    appendix="${target##/}"
    relative=''
    while appendix="${target#"$current"/}"
        [ "$current" != '/' ] && [ "$appendix" = "$target" ]; do
        if [ "$current" = "$appendix" ]; then
            relative="${relative:-.}"
            echo "${relative#/}"
            return 0
        fi
        current="${current%/*}"
        relative="$relative${relative:+/}.."
    done
    relative="$relative${relative:+${appendix:+/}}${appendix#/}"
    echo "$relative"
}
relpath "$@"

ZSH壳函数

现在是更加健壮的zsh版本。如果您希望它将参数解析为真实路径à la realpath -f(在Linux coreutils包中可用),请将第3行和第4行上的:a替换为:a。

要在zsh中使用它,请删除第一行和最后一行,并将其放在$FPATH变量中的目录中。

#!/usr/bin/env zsh
relpath () {
    [[ $# -ge 1 ]] && [[ $# -le 2 ]] || return 1
    local target=${${2:-$1}:a} # replace `:a' by `:A` to resolve symlinks
    local current=${${${2:+$1}:-$PWD}:a} # replace `:a' by `:A` to resolve symlinks
    local appendix=${target#/}
    local relative=''
    while appendix=${target#$current/}
        [[ $current != '/' ]] && [[ $appendix = $target ]]; do
        if [[ $current = $appendix ]]; then
            relative=${relative:-.}
            print ${relative#/}
            return 0
        fi
        current=${current%/*}
        relative="$relative${relative:+/}.."
    done
    relative+=${relative:+${appendix:+/}}${appendix#/}
    print $relative
}
relpath "$@"

测试脚本

最后是测试脚本。它接受一个选项,即-v来启用详细输出。

#!/usr/bin/env zsh
set -eu
VERBOSE=false
script_name=$(basename $0)

usage () {
    print "\n    Usage: $script_name SRC_PATH DESTINATION_PATH\n" >&2
    exit ${1:=1}
}
vrb () { $VERBOSE && print -P ${(%)@} || return 0; }

relpath_check () {
    [[ $# -ge 1 ]] && [[ $# -le 2 ]] || return 1
    target=${${2:-$1}}
    prefix=${${${2:+$1}:-$PWD}}
    result=$(relpath $prefix $target)
    # Compare with python's os.path.relpath function
    py_result=$(python -c "import os.path; print os.path.relpath('$target', '$prefix')")
    col='%F{green}'
    if [[ $result != $py_result ]] && col='%F{red}' || $VERBOSE; then
        print -P "${col}Source: '$prefix'\nDestination: '$target'%f"
        print -P "${col}relpath: ${(qq)result}%f"
        print -P "${col}python:  ${(qq)py_result}%f\n"
    fi
}

run_checks () {
    print "Running checks..."

    relpath_check '/    a   b/å/⮀*/!' '/    a   b/å/⮀/xäå/?'

    relpath_check '/'  '/A'
    relpath_check '/A'  '/'
    relpath_check '/  & /  !/*/\\/E' '/'
    relpath_check '/' '/  & /  !/*/\\/E'
    relpath_check '/  & /  !/*/\\/E' '/  & /  !/?/\\/E/F'
    relpath_check '/X/Y' '/  & /  !/C/\\/E/F'
    relpath_check '/  & /  !/C' '/A'
    relpath_check '/A /  !/C' '/A /B'
    relpath_check '/Â/  !/C' '/Â/  !/C'
    relpath_check '/  & /B / C' '/  & /B / C/D'
    relpath_check '/  & /  !/C' '/  & /  !/C/\\/Ê'
    relpath_check '/Å/  !/C' '/Å/  !/D'
    relpath_check '/.A /*B/C' '/.A /*B/\\/E'
    relpath_check '/  & /  !/C' '/  & /D'
    relpath_check '/  & /  !/C' '/  & /\\/E'
    relpath_check '/  & /  !/C' '/\\/E/F'

    relpath_check /home/part1/part2 /home/part1/part3
    relpath_check /home/part1/part2 /home/part4/part5
    relpath_check /home/part1/part2 /work/part6/part7
    relpath_check /home/part1       /work/part1/part2/part3/part4
    relpath_check /home             /work/part2/part3
    relpath_check /                 /work/part2/part3/part4
    relpath_check /home/part1/part2 /home/part1/part2/part3/part4
    relpath_check /home/part1/part2 /home/part1/part2/part3
    relpath_check /home/part1/part2 /home/part1/part2
    relpath_check /home/part1/part2 /home/part1
    relpath_check /home/part1/part2 /home
    relpath_check /home/part1/part2 /
    relpath_check /home/part1/part2 /work
    relpath_check /home/part1/part2 /work/part1
    relpath_check /home/part1/part2 /work/part1/part2
    relpath_check /home/part1/part2 /work/part1/part2/part3
    relpath_check /home/part1/part2 /work/part1/part2/part3/part4 
    relpath_check home/part1/part2 home/part1/part3
    relpath_check home/part1/part2 home/part4/part5
    relpath_check home/part1/part2 work/part6/part7
    relpath_check home/part1       work/part1/part2/part3/part4
    relpath_check home             work/part2/part3
    relpath_check .                work/part2/part3
    relpath_check home/part1/part2 home/part1/part2/part3/part4
    relpath_check home/part1/part2 home/part1/part2/part3
    relpath_check home/part1/part2 home/part1/part2
    relpath_check home/part1/part2 home/part1
    relpath_check home/part1/part2 home
    relpath_check home/part1/part2 .
    relpath_check home/part1/part2 work
    relpath_check home/part1/part2 work/part1
    relpath_check home/part1/part2 work/part1/part2
    relpath_check home/part1/part2 work/part1/part2/part3
    relpath_check home/part1/part2 work/part1/part2/part3/part4

    print "Done with checks."
}
if [[ $# -gt 0 ]] && [[ $1 = "-v" ]]; then
    VERBOSE=true
    shift
fi
if [[ $# -eq 0 ]]; then
    run_checks
else
    VERBOSE=true
    relpath_check "$@"
fi

我需要这样的东西,但它也解决了符号链接。我发现pwd有一个-P标志用于此目的。附加了我的脚本的一个片段。它在shell脚本的函数中,因此是$1和$2。结果值是从START_ABS到END_ABS的相对路径,位于UPDIRS变量中。为了执行pwd -P,将脚本cd放入每个参数目录,这也意味着将处理相对路径参数。干杯,吉姆

SAVE_DIR="$PWD"
cd "$1"
START_ABS=`pwd -P`
cd "$SAVE_DIR"
cd "$2"
END_ABS=`pwd -P`

START_WORK="$START_ABS"
UPDIRS=""

while test -n "${START_WORK}" -a "${END_ABS/#${START_WORK}}" '==' "$END_ABS";
do
    START_WORK=`dirname "$START_WORK"`"/"
    UPDIRS=${UPDIRS}"../"
done
UPDIRS="$UPDIRS${END_ABS/#${START_WORK}}"
cd "$SAVE_DIR"

我使用的macOS默认情况下没有realpath命令,所以我做了一个纯bash函数来计算它。

#!/bin/bash

##
# print a relative path from "source folder" to "target file"
#
# params:
#  $1 - target file, can be a relative path or an absolute path.
#  $2 - source folder, can be a relative path or an absolute path.
#
# test:
#  $ mkdir -p ~/A/B/C/D; touch ~/A/B/C/D/testfile.txt; touch ~/A/B/testfile.txt
#
#  $ getRelativePath ~/A/B/C/D/testfile.txt  ~/A/B
#  $ C/D/testfile.txt
#  
#  $ getRelativePath ~/A/B/testfile.txt  ~/A/B/C
#  $ ../testfile.txt
#
#  $ getRelativePath ~/A/B/testfile.txt  /
#  $ home/bunnier/A/B/testfile.txt 
#
function getRelativePath(){
    local targetFilename=$(basename $1)
    local targetFolder=$(cd $(dirname $1);pwd) # absolute target folder path
    local currentFolder=$(cd $2;pwd) # absulute source folder
    local result=.

    while [ "$currentFolder" != "$targetFolder" ];do
      if [[ "$targetFolder" =~ "$currentFolder"* ]];then
          pointSegment=${targetFolder#$currentFolder}
          result=$result/${pointSegment#/}
          break
      fi  
      result="$result"/..
      currentFolder=$(dirname $currentFolder)
    done

    result=$result/$targetFilename
    echo ${result#./}
}

这个脚本只对路径名有效。它不需要任何文件存在。如果传递的路径不是绝对的,那么行为就有点不寻常,但是如果两条路径都是相对的,那么应该能正常工作。

我只在OS X上测试过,所以可能不太便携。

#!/bin/bash
set -e
declare SCRIPT_NAME="$(basename $0)"
function usage {
    echo "Usage: $SCRIPT_NAME <base path> <target file>"
    echo "       Outputs <target file> relative to <base path>"
    exit 1
}

if [ $# -lt 2 ]; then usage; fi

declare base=$1
declare target=$2
declare -a base_part=()
declare -a target_part=()

#Split path elements & canonicalize
OFS="$IFS"; IFS='/'
bpl=0;
for bp in $base; do
    case "$bp" in
        ".");;
        "..") let "bpl=$bpl-1" ;;
        *) base_part[${bpl}]="$bp" ; let "bpl=$bpl+1";;
    esac
done
tpl=0;
for tp in $target; do
    case "$tp" in
        ".");;
        "..") let "tpl=$tpl-1" ;;
        *) target_part[${tpl}]="$tp" ; let "tpl=$tpl+1";;
    esac
done
IFS="$OFS"

#Count common prefix
common=0
for (( i=0 ; i<$bpl ; i++ )); do
    if [ "${base_part[$i]}" = "${target_part[$common]}" ] ; then
        let "common=$common+1"
    else
        break
    fi
done

#Compute number of directories up
let "updir=$bpl-$common" || updir=0 #if the expression is zero, 'let' fails

#trivial case (after canonical decomposition)
if [ $updir -eq 0 ]; then
    echo .
    exit
fi

#Print updirs
for (( i=0 ; i<$updir ; i++ )); do
    echo -n ../
done

#Print remaining path
for (( i=$common ; i<$tpl ; i++ )); do
    if [ $i -ne $common ]; then
        echo -n "/"
    fi
    if [ "" != "${target_part[$i]}" ] ; then
        echo -n "${target_part[$i]}"
    fi
done
#One last newline
echo